Question

If $f : R \rightarrow[-1,2], f ( x )=\frac{ x ^2+9 bx +17}{ ax ^3+ x ^2+ bx +33}$ is onto function and $f ^{\prime}( d )= f ^{\prime}( e )=0$. Then

• A. eccentricity of curve $\frac{x^2}{(d+e)^2}+\frac{y^2}{b+1}=1$ can be $\frac{1}{2}$
• B. eccentricity of curve $\frac{x^2}{(d+e)^2}+\frac{y^2}{b+1}=1$ can be $\frac{\sqrt{5}}{2}$
• C. eccentricity of curve $\frac{x^2}{(d+e)^2}+\frac{y^2}{a+1}=1$ can be $\frac{\sqrt{3}}{2}$
• D. eccentricity of curve $\frac{x^2}{(d+e)^2}+\frac{y^2}{a+1}=1$ can be $\frac{\sqrt{7}}{2}$.

Answer 1

Answer: (A), (B), (C)

$ \because$ Domain of function is real number.
$\therefore a =0$
Now, $y=\frac{x^2+9 b x+17}{x^2+b x+33} \Rightarrow(y-1) x^2+(b y-9 b) x+(33 y-17)=0$
$\because x$ is real
$\therefore(b y-9 b)^2-4(y-1)(33 y-17) \geq 0$
$\because$ Range is $[-1,2]$
$\therefore-1$ and 2 should be root
$\therefore b =+2$ and -2
$\therefore$ for $b=2 \Rightarrow d=-5, e=7$
and $b=-2 \Rightarrow d=-7, e=+5$